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Re: Can I get ComplexExpand to really work?

Two observations: (1) on assumptions in Integrate(2) on your sequence.

(1) The Integral

As has already been pointed out by several people

E^(I a b x^2),{x, -Infinity, Infinity}, Assumptions -> {a>0, b>0}

If[Im[a*b] == 0, (Sqrt[Pi/2]*(1 + I*Sign[a*b]))/
   (Sqrt[a]*Sqrt[b]), Integrate[E^(I*a*b*x^2),
   {x, -Infinity, Infinity}]]

gives nearly what you want

If we help Mathematica a bit we get

E^(I a b x^2),{x, -Infinity, Infinity}, Assumptions -> {a b>0,a>0,b>0}

((1 + I)*Sqrt[Pi/2])/(Sqrt[a]*Sqrt[b])

Which, I guess, points up the need to internally extend the assumptions
in the manner of the package Declare.

(2) The sequence

The sequence you gave was
data = { {1, 0}, {2, 1/4}, {3, 1}, {4, 5/2}, {5, 5}, {6, 35/4}, {7, 14},
      {8, 21}, {9, 30}, {10, 165/4}, {11, 55}}; and your answer, given
later,  was that this is given by the function

f(N) = (N-1) N (N+1) / 24

Here is a Mathematica deduction of this

p[n_] := InterpolatingPolynomial[Take[data, n], x]

Look for where p[n] repeats

fp = (n=1;FixedPoint[(++n; p[n]) &, 1])

(1/4 + (1/4 + 1/24*(-3 + x))*(-2 + x))*(-1 + x)

f[x_] = Simplify[fp]

1/24*(-1 + x)*x*(1 + x)


Last/@data == Table[f[n], {n, Length[data]}]


In a symbolic case, having used Mathematica to guess the formula, we
might use induction to prove the result.

Allan Hayes
Mathematica Training and Consulting
hay at
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

Topher Cawlfield wrote in message <71bkvu$pul at>...
>Many thanks for the help!
>(I'm not quite sure how to best respond since this newsgroup seems to be
>acting funny)
>Daniel Lichtblau, P.J. Hinton, and Lars Hohmuth suggested that I use the
>"Assumptions" option with Integrate.  Thanks, I didn't know about that!
>This works well in the most simple case, but doesn't work quite right
>in more general cases.
>Integrate[E^(I a b x^2), {x, -Infinity, Infinity}, Assumptions->{a>0 ,
>Essentially returns Sqrt[Pi/(2 a b)] (1 + I Sign[a b])
>This is progress, but I could make the argument that if a>0 and b>0,
>then  Sign[a b] = 1.
>If I use "Assumptions->{a b > 0}" instead, I get the more simple result.
>M. R. Schaferkotter suggested that I try the Declare.m package from
>mathsource.  I haven't tried this yet, but I will soon.
>For the specific problem I was working on, I was finally able to use
>Mathematica very successfully!  I thought I would post it here since it
>worked so well, and must have saved me at least a month of grinding
>through algebra and making mistakes.  Well, it didn't save me a month
>really, it just made one technique practical whereas it wouldn't have
>been otherwise.
>I want to use a Feynman path integral approach to find the transition
>amplitude for an electron in a constant electric field.  This boils
>down to doing a lot of integrals, noticing a pattern, and deducing the
>limiting case of performing an infinite number of integrals.
>I won't go into all the details, but I would end up needing to perform
>integrals like:
>Integrate[E^(I(a x^2 + b x y + c y^2 + d y z + e z^2 + f)),
> {x, -Infinity, Infinity}, {y, -Infinity, Infinity}, {z, -Infinity,
>where a, b, c, d, e, and f are simple ratios of physical constants and
>parameters in the problem.
>The best way to evaluate these by hand is to first "complete the square"
>in x:
>Integrate[E^(I(a (x+junk)^2 + more junk)),...]
>The value of "junk" doesn't matter since you can rescale x.  The limits
>of integration, being infinite, don't change.  You can use the
>integration rule I was getting at above.  Then, repeat the process by
>completing the square in y of "more junk", applying the integration
>rule again, and moving on to z.
>I couldn't find any functions in mathematica that would complete the
>square for me, but the CoefficientList function helped a lot.  I made
>my own function:
>CompleteTheSquare[p_, x_]  := Module[{coeffs, fp},
> coeffs = CoefficientList[p, x];
> fp = coeffs[[3]] (x + coeffs[[2]] / (2 coeffs[[3]]) )^2;
> parts = {coeffs[[3]], Simplify[p - fp], fp};
> fp + parts[[2]]]  ]
>This function returns the modified polynomial, and also sets the symbol
>'parts' to a list containing the various pieces of the result.
>I then defined a function that "did the integration": MyInt[a_] :=
>Sqrt[- Pi / a]
>So that the integral of E^(a (x-b)^2 + c) is MyInt[a] E^c.
>And finally I made a function that did a Do loop over the integrals.
>This was a bit of work, but the results were really quite nice.  By
>hand I could have only done 1 or 2 such integrals before tiring, and
>that would have taken me hours.  With this I could try any number.  I
>went up to 10, which just took a Mathematica a couple minutes.
>The morals of the story (IMO):  (1) Use Mathematica for what it's good
>at.  (2) One still needs to be somewhat knowledgable of and practiced
>with mathematical techniques.  Even though Mathematica can do integrals
>fairly well, it's still important to know how to do them "by hand".
>This is a good thing.
>- Topher Cawlfield
>p.s.  The fun part of the problem, besides the Mathematica programming,
>was figuring out a strange number series that came up:
>What's f(N)?
>N  | f(N)
>1  | 0
>2  | 1/4
>3  | 1
>4  | 5/2
>5  | 5
>6  | 35/4
>7  | 14
>8  | 21
>9  | 30
>10 | 165/4
>11 | 55
>I'll post the answer tomorrow.  These were coefficients in the answer.
>To get one of them requires doing N-1 integrals symbolicly.

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